Direct passage from $n$ prime to $n$ non-prime in Euler Totient Function $Phi$












0












$begingroup$


I wish to derive, for a proof, the Euler Totient Function starting from the case $n$ prime to $n$ non-prime.



Let $n$ prime, we know $Phi(n)=n-1$. But what if now I assume $n=p_0^{a_0}p_1^{a_1}...p_r^{a_r}$ and I want to reach $Phi(n)=n prod_{p|n}^{}(1-1/p) $ from there? Like:



$Phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})=p_0^{a_0}p_1^{a_1}...p_r^{a_r}-1 Rightarrow p_0^{a_0}p_1^{a_1}...p_r^{a_r}(1-1/(p_0^{a_0}p_1^{a_1}...p_r^{a_r}))=?$



Is this kind of approach possible for this formula?






abstract-algebra elementary-number-theory modular-arithmetic totient-function







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    The two steps are $phi(p^k) = p^k-p^{k-1}$ and $phi(nm) = phi(n) phi(m)$ whenever $gcd(n,m)=1$ thus $phi(n) = prod_{p^k |n} phi(p^k) = nprod_{p^k |n} frac{phi(p^k) }{p^k}= n prod_{p| n} (1-p^{-1})$
    $endgroup$
    – reuns
    Jan 15 at 8:19








  • 1




    $begingroup$
    In particular, it is not true that $phi(p_0^{a_0}p_1^{a_1}cdots p_r^{a_r})=p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}-1$ when $p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}$ is not prime.
    $endgroup$
    – Greg Martin
    Jan 15 at 8:28










  • $begingroup$
    So such kind of derivation, as I suppose, it's not possible. Thanks
    $endgroup$
    – Alessar
    Jan 15 at 8:31










  • $begingroup$
    @reuns the symbol "$||$" means for "that divide"?
    $endgroup$
    – Alessar
    Jan 15 at 9:13






  • 1




    $begingroup$
    $p^k| n$ iff $p^k | n$ and $p^{k+1} nmid n$. In this setting $p$ means (ranging over) the primes and $p^k$ means over the prime powers
    $endgroup$
    – reuns
    Jan 15 at 9:17


















0












$begingroup$


I wish to derive, for a proof, the Euler Totient Function starting from the case $n$ prime to $n$ non-prime.



Let $n$ prime, we know $Phi(n)=n-1$. But what if now I assume $n=p_0^{a_0}p_1^{a_1}...p_r^{a_r}$ and I want to reach $Phi(n)=n prod_{p|n}^{}(1-1/p) $ from there? Like:



$Phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})=p_0^{a_0}p_1^{a_1}...p_r^{a_r}-1 Rightarrow p_0^{a_0}p_1^{a_1}...p_r^{a_r}(1-1/(p_0^{a_0}p_1^{a_1}...p_r^{a_r}))=?$



Is this kind of approach possible for this formula?






abstract-algebra elementary-number-theory modular-arithmetic totient-function







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    The two steps are $phi(p^k) = p^k-p^{k-1}$ and $phi(nm) = phi(n) phi(m)$ whenever $gcd(n,m)=1$ thus $phi(n) = prod_{p^k |n} phi(p^k) = nprod_{p^k |n} frac{phi(p^k) }{p^k}= n prod_{p| n} (1-p^{-1})$
    $endgroup$
    – reuns
    Jan 15 at 8:19








  • 1




    $begingroup$
    In particular, it is not true that $phi(p_0^{a_0}p_1^{a_1}cdots p_r^{a_r})=p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}-1$ when $p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}$ is not prime.
    $endgroup$
    – Greg Martin
    Jan 15 at 8:28










  • $begingroup$
    So such kind of derivation, as I suppose, it's not possible. Thanks
    $endgroup$
    – Alessar
    Jan 15 at 8:31










  • $begingroup$
    @reuns the symbol "$||$" means for "that divide"?
    $endgroup$
    – Alessar
    Jan 15 at 9:13






  • 1




    $begingroup$
    $p^k| n$ iff $p^k | n$ and $p^{k+1} nmid n$. In this setting $p$ means (ranging over) the primes and $p^k$ means over the prime powers
    $endgroup$
    – reuns
    Jan 15 at 9:17
















0












0








0





$begingroup$


I wish to derive, for a proof, the Euler Totient Function starting from the case $n$ prime to $n$ non-prime.



Let $n$ prime, we know $Phi(n)=n-1$. But what if now I assume $n=p_0^{a_0}p_1^{a_1}...p_r^{a_r}$ and I want to reach $Phi(n)=n prod_{p|n}^{}(1-1/p) $ from there? Like:



$Phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})=p_0^{a_0}p_1^{a_1}...p_r^{a_r}-1 Rightarrow p_0^{a_0}p_1^{a_1}...p_r^{a_r}(1-1/(p_0^{a_0}p_1^{a_1}...p_r^{a_r}))=?$



Is this kind of approach possible for this formula?






abstract-algebra elementary-number-theory modular-arithmetic totient-function







share|cite|improve this question









$endgroup$




I wish to derive, for a proof, the Euler Totient Function starting from the case $n$ prime to $n$ non-prime.



Let $n$ prime, we know $Phi(n)=n-1$. But what if now I assume $n=p_0^{a_0}p_1^{a_1}...p_r^{a_r}$ and I want to reach $Phi(n)=n prod_{p|n}^{}(1-1/p) $ from there? Like:



$Phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})=p_0^{a_0}p_1^{a_1}...p_r^{a_r}-1 Rightarrow p_0^{a_0}p_1^{a_1}...p_r^{a_r}(1-1/(p_0^{a_0}p_1^{a_1}...p_r^{a_r}))=?$



Is this kind of approach possible for this formula?






abstract-algebra elementary-number-theory modular-arithmetic totient-function




abstract-algebra elementary-number-theory modular-arithmetic totient-function






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 15 at 8:17









AlessarAlessar

29615




29615








  • 3




    $begingroup$
    The two steps are $phi(p^k) = p^k-p^{k-1}$ and $phi(nm) = phi(n) phi(m)$ whenever $gcd(n,m)=1$ thus $phi(n) = prod_{p^k |n} phi(p^k) = nprod_{p^k |n} frac{phi(p^k) }{p^k}= n prod_{p| n} (1-p^{-1})$
    $endgroup$
    – reuns
    Jan 15 at 8:19








  • 1




    $begingroup$
    In particular, it is not true that $phi(p_0^{a_0}p_1^{a_1}cdots p_r^{a_r})=p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}-1$ when $p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}$ is not prime.
    $endgroup$
    – Greg Martin
    Jan 15 at 8:28










  • $begingroup$
    So such kind of derivation, as I suppose, it's not possible. Thanks
    $endgroup$
    – Alessar
    Jan 15 at 8:31










  • $begingroup$
    @reuns the symbol "$||$" means for "that divide"?
    $endgroup$
    – Alessar
    Jan 15 at 9:13






  • 1




    $begingroup$
    $p^k| n$ iff $p^k | n$ and $p^{k+1} nmid n$. In this setting $p$ means (ranging over) the primes and $p^k$ means over the prime powers
    $endgroup$
    – reuns
    Jan 15 at 9:17
















  • 3




    $begingroup$
    The two steps are $phi(p^k) = p^k-p^{k-1}$ and $phi(nm) = phi(n) phi(m)$ whenever $gcd(n,m)=1$ thus $phi(n) = prod_{p^k |n} phi(p^k) = nprod_{p^k |n} frac{phi(p^k) }{p^k}= n prod_{p| n} (1-p^{-1})$
    $endgroup$
    – reuns
    Jan 15 at 8:19








  • 1




    $begingroup$
    In particular, it is not true that $phi(p_0^{a_0}p_1^{a_1}cdots p_r^{a_r})=p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}-1$ when $p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}$ is not prime.
    $endgroup$
    – Greg Martin
    Jan 15 at 8:28










  • $begingroup$
    So such kind of derivation, as I suppose, it's not possible. Thanks
    $endgroup$
    – Alessar
    Jan 15 at 8:31










  • $begingroup$
    @reuns the symbol "$||$" means for "that divide"?
    $endgroup$
    – Alessar
    Jan 15 at 9:13






  • 1




    $begingroup$
    $p^k| n$ iff $p^k | n$ and $p^{k+1} nmid n$. In this setting $p$ means (ranging over) the primes and $p^k$ means over the prime powers
    $endgroup$
    – reuns
    Jan 15 at 9:17










3




3




$begingroup$
The two steps are $phi(p^k) = p^k-p^{k-1}$ and $phi(nm) = phi(n) phi(m)$ whenever $gcd(n,m)=1$ thus $phi(n) = prod_{p^k |n} phi(p^k) = nprod_{p^k |n} frac{phi(p^k) }{p^k}= n prod_{p| n} (1-p^{-1})$
$endgroup$
– reuns
Jan 15 at 8:19






$begingroup$
The two steps are $phi(p^k) = p^k-p^{k-1}$ and $phi(nm) = phi(n) phi(m)$ whenever $gcd(n,m)=1$ thus $phi(n) = prod_{p^k |n} phi(p^k) = nprod_{p^k |n} frac{phi(p^k) }{p^k}= n prod_{p| n} (1-p^{-1})$
$endgroup$
– reuns
Jan 15 at 8:19






1




1




$begingroup$
In particular, it is not true that $phi(p_0^{a_0}p_1^{a_1}cdots p_r^{a_r})=p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}-1$ when $p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}$ is not prime.
$endgroup$
– Greg Martin
Jan 15 at 8:28




$begingroup$
In particular, it is not true that $phi(p_0^{a_0}p_1^{a_1}cdots p_r^{a_r})=p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}-1$ when $p_0^{a_0}p_1^{a_1}cdots p_r^{a_r}$ is not prime.
$endgroup$
– Greg Martin
Jan 15 at 8:28












$begingroup$
So such kind of derivation, as I suppose, it's not possible. Thanks
$endgroup$
– Alessar
Jan 15 at 8:31




$begingroup$
So such kind of derivation, as I suppose, it's not possible. Thanks
$endgroup$
– Alessar
Jan 15 at 8:31












$begingroup$
@reuns the symbol "$||$" means for "that divide"?
$endgroup$
– Alessar
Jan 15 at 9:13




$begingroup$
@reuns the symbol "$||$" means for "that divide"?
$endgroup$
– Alessar
Jan 15 at 9:13




1




1




$begingroup$
$p^k| n$ iff $p^k | n$ and $p^{k+1} nmid n$. In this setting $p$ means (ranging over) the primes and $p^k$ means over the prime powers
$endgroup$
– reuns
Jan 15 at 9:17






$begingroup$
$p^k| n$ iff $p^k | n$ and $p^{k+1} nmid n$. In this setting $p$ means (ranging over) the primes and $p^k$ means over the prime powers
$endgroup$
– reuns
Jan 15 at 9:17












1 Answer
1






active

oldest

votes


















1












$begingroup$

To be precise, your "kind" of derivation is not possible. This is because $phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})neq p_0^{a_0}p_1^{a_1}...p_r^{a_r}-1$.



Think about what the totient function entails: the number of integers less than the input that are coprime with it. Only with prime inputs is this one less than the input.



The totient function, however, does retain two properties, namely as reuns mentioned, that $phi(p^k)=p^k-p^{k-1}$ and $phi(n)phi(m)=phi(mn)$ when $m$ and $n$ are coprime. The proof of these is easily verifiable given a google search.



Thus, we can have a "direct" approach. Since powers of primes are always coprime to one another,
begin{align*}
phi(n)&=phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})\
&=phi(p_0^{a_0})phi(p_1^{a_1})...phi(p_r^{a_r})\
&=(p_0^{a_0}-p_0^{a_0-1})(p_1^{a_1}-p_1^{a_1-1})...(p_r^{a_r}-p_r^{a_r-1})\
&=p_0^{a_0}(1-1/p_0)p_1^{a_1}(1-1/p_1)...p_r^{a_r}(1-1/p_r)\
&=ndisplaystyleprod_{p|n}(1-1/p).
end{align*}






share|cite|improve this answer









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    $begingroup$

    To be precise, your "kind" of derivation is not possible. This is because $phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})neq p_0^{a_0}p_1^{a_1}...p_r^{a_r}-1$.



    Think about what the totient function entails: the number of integers less than the input that are coprime with it. Only with prime inputs is this one less than the input.



    The totient function, however, does retain two properties, namely as reuns mentioned, that $phi(p^k)=p^k-p^{k-1}$ and $phi(n)phi(m)=phi(mn)$ when $m$ and $n$ are coprime. The proof of these is easily verifiable given a google search.



    Thus, we can have a "direct" approach. Since powers of primes are always coprime to one another,
    begin{align*}
    phi(n)&=phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})\
    &=phi(p_0^{a_0})phi(p_1^{a_1})...phi(p_r^{a_r})\
    &=(p_0^{a_0}-p_0^{a_0-1})(p_1^{a_1}-p_1^{a_1-1})...(p_r^{a_r}-p_r^{a_r-1})\
    &=p_0^{a_0}(1-1/p_0)p_1^{a_1}(1-1/p_1)...p_r^{a_r}(1-1/p_r)\
    &=ndisplaystyleprod_{p|n}(1-1/p).
    end{align*}






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      To be precise, your "kind" of derivation is not possible. This is because $phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})neq p_0^{a_0}p_1^{a_1}...p_r^{a_r}-1$.



      Think about what the totient function entails: the number of integers less than the input that are coprime with it. Only with prime inputs is this one less than the input.



      The totient function, however, does retain two properties, namely as reuns mentioned, that $phi(p^k)=p^k-p^{k-1}$ and $phi(n)phi(m)=phi(mn)$ when $m$ and $n$ are coprime. The proof of these is easily verifiable given a google search.



      Thus, we can have a "direct" approach. Since powers of primes are always coprime to one another,
      begin{align*}
      phi(n)&=phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})\
      &=phi(p_0^{a_0})phi(p_1^{a_1})...phi(p_r^{a_r})\
      &=(p_0^{a_0}-p_0^{a_0-1})(p_1^{a_1}-p_1^{a_1-1})...(p_r^{a_r}-p_r^{a_r-1})\
      &=p_0^{a_0}(1-1/p_0)p_1^{a_1}(1-1/p_1)...p_r^{a_r}(1-1/p_r)\
      &=ndisplaystyleprod_{p|n}(1-1/p).
      end{align*}






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        To be precise, your "kind" of derivation is not possible. This is because $phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})neq p_0^{a_0}p_1^{a_1}...p_r^{a_r}-1$.



        Think about what the totient function entails: the number of integers less than the input that are coprime with it. Only with prime inputs is this one less than the input.



        The totient function, however, does retain two properties, namely as reuns mentioned, that $phi(p^k)=p^k-p^{k-1}$ and $phi(n)phi(m)=phi(mn)$ when $m$ and $n$ are coprime. The proof of these is easily verifiable given a google search.



        Thus, we can have a "direct" approach. Since powers of primes are always coprime to one another,
        begin{align*}
        phi(n)&=phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})\
        &=phi(p_0^{a_0})phi(p_1^{a_1})...phi(p_r^{a_r})\
        &=(p_0^{a_0}-p_0^{a_0-1})(p_1^{a_1}-p_1^{a_1-1})...(p_r^{a_r}-p_r^{a_r-1})\
        &=p_0^{a_0}(1-1/p_0)p_1^{a_1}(1-1/p_1)...p_r^{a_r}(1-1/p_r)\
        &=ndisplaystyleprod_{p|n}(1-1/p).
        end{align*}






        share|cite|improve this answer









        $endgroup$



        To be precise, your "kind" of derivation is not possible. This is because $phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})neq p_0^{a_0}p_1^{a_1}...p_r^{a_r}-1$.



        Think about what the totient function entails: the number of integers less than the input that are coprime with it. Only with prime inputs is this one less than the input.



        The totient function, however, does retain two properties, namely as reuns mentioned, that $phi(p^k)=p^k-p^{k-1}$ and $phi(n)phi(m)=phi(mn)$ when $m$ and $n$ are coprime. The proof of these is easily verifiable given a google search.



        Thus, we can have a "direct" approach. Since powers of primes are always coprime to one another,
        begin{align*}
        phi(n)&=phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})\
        &=phi(p_0^{a_0})phi(p_1^{a_1})...phi(p_r^{a_r})\
        &=(p_0^{a_0}-p_0^{a_0-1})(p_1^{a_1}-p_1^{a_1-1})...(p_r^{a_r}-p_r^{a_r-1})\
        &=p_0^{a_0}(1-1/p_0)p_1^{a_1}(1-1/p_1)...p_r^{a_r}(1-1/p_r)\
        &=ndisplaystyleprod_{p|n}(1-1/p).
        end{align*}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 15 at 20:50









        Tejas RaoTejas Rao

        16611




        16611






























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            Dobbiaco

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            Dobbiaco comune ( IT ) Dobbiaco ( DE ) Toblach Dobbiaco Vecchia Localizzazione Stato   Italia Regione Trentino-Alto Adige Provincia Bolzano Amministrazione Sindaco Guido Bocher (lista civica Indipendenti-Unabhängige) dal 10-5-2015 (2º mandato) Lingue ufficiali Tedesco, Italiano Territorio Coordinate 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Coordinate: 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Altitudine 1 213 m s.l.m. Superficie 125,42 km² Abitanti 3 356 [2] (31-7-2017) Densità 26,76 ab./km² Frazioni Carbonin ( Schluderbach ), Santa Maria ( Aufkirchen ), Valle San Silvestro ( Wahlen ) Comuni confinanti Auronzo di Cadore (BL), Braies, Cortina d'Ampezzo (BL), Innervillgraten (AT-7), San Candido, Sesto, Valle di Casies, Villabassa Altre informazioni Cod. postale 39034 Prefisso 0474 Fuso...
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